我不明白为什么 string.size 返回它的作用答案

【问题标题】：I don't understand why string.size returns what it does我不明白为什么 string.size 返回它的作用
【发布时间】：2011-09-14 07:58:06
【问题描述】：

long_string = <<EOS
It was the best of times,
It was the worst of times.
EOS

返回 53。为什么？空格算不算？即便如此。我们如何得到 53？

这个怎么样？

     def test_flexible_quotes_can_handle_multiple_lines
    long_string = %{
It was the best of times,
It was the worst of times.
}
    assert_equal 54, long_string.size
  end

  def test_here_documents_can_also_handle_multiple_lines
    long_string = <<EOS
It was the best of times,
It was the worst of times.
EOS
    assert_equal 53, long_string.size
  end

是这样吗，因为 %{ 案例将每个 /n 视为一个字符，并且在第一行之前被认为是一个字符，在末尾一个，然后在第二行末尾，而在 @ 987654324@ 情况下只有一个在第一行之前和一个在第一行之后？也就是说，为什么前者是54，后者是53？

【问题讨论】：

哦，请不要引用查尔斯狄更斯的名言......

标签： ruby heredoc

【解决方案1】：

为：

long_string = <<EOS
It was the best of times,
It was the worst of times.
EOS

String is:
"It was the best of times,\nIt was the worst of times.\n"

It was the best of times, => 25
<newline> => 1
It was the worst of times. => 26
<newline> => 1
Total = 25 + 1 + 26 + 1 = 53

和

long_string = %{
It was the best of times,
It was the worst of times.
}

String is:
"\nIt was the best of times,\nIt was the worst of times.\n"
#Note leading "\n"

工作原理：

对于<<EOS，它后面的行是字符串的一部分。 << 之后的所有文本与 << 位于同一行并到行尾将是确定字符串何时结束的“标记”的一部分（在这种情况下，一行上的 EOS 本身是匹配<<EOS)。

在%{...} 的情况下，它只是用于代替"..." 的不同分隔符。因此，当字符串在 %{ 之后的新行开始时，该换行符就是字符串的一部分。

试试这个例子，你会看到%{...} 和"..." 的工作原理是一样的：

a = "
It was the best of times,
It was the worst of times.
"
a.length # => 54

b = "It was the best of times,
It was the worst of times.
"
b.length # => 53

【讨论】：